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Math Help - [SOLVED] Intgers

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    [SOLVED] Intgers

    If a and b are positive integers and (a^{1/2}b^{1/3})^6=432, what is the value of ab?

    Can you please show steps?

    Thanks in advance

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    If a and b are positive integers and (a^{1/2}b^{1/3})^6=432, what is the value of ab?

    Can you please show steps?

    Thanks in advance

    - - - - - 596
    note that you have a^3b^2 = 432

    so you want to express 432 as the product of a square and a cube. find the prime factorization of 432, that should help
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    Quote Originally Posted by Jhevon View Post
    note that you have a^3b^2 = 432

    so you want to express 432 as the product of a square and a cube. find the prime factorization of 432, that should help
    you mean 3^3 x  2^4?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    you mean 3^3 x  2^4?
    yes

    can you express that as something cubed times something squared?
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    How about 4^2 times 3^3? it's something squared and something cubed
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    How about 4^2 times 3^3? it's something squared and something cubed
    yes. so what is your a and b? and hence, ab
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    thank you. so the final answer should be 4x3=12

    I have a question from your earlier post:

    Quote Originally Posted by Jhevon View Post
    note that you have a^3b^2 = 432
    how did you get from a^\frac{1}{2}b^\frac{1}{3}=432 to a^3b^2 = 432?? and also what about the 6 from [a^(1/2)b^(1/3)]^6? How about the 6 here? Thanks again!!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by fabxx View Post
    thank you. so the final answer should be 4x3=12
    yes

    I have a question from your earlier post:


    how did you get from a^\frac{1}{2}b^\frac{1}{3}=432 to a^3b^2 = 432?? and also what about the 6 from [a^(1/2)b^(1/3)]^6? How about the 6 here? Thanks again!!
    here is the rule: (x^a)^b = x^{ab} and also (x^ay^b)^c = x^{ac}y^{bc}

    that is, we can distribute the power and multiply them when we raise a number to a power to another power
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